The port of South Louisiana, located along miles of the Mississippi River between New Orleans and Baton Rouge, is the largest bulk cargo port in the world. The U. S. Army Corps of Engineers reports that the port handles a mean of million tons of cargo per week (USA Today, September). Assume that the number of tons of cargo handled per week is normally distributed with a standard deviation of million tons.
A. What is the probability that the port handles less than 5 million tons of cargo per week?
B. What is the probability that the port handles 3 or more million tons of cargo per week?
C. What is the probability that the port handles between 3 million and 4 million tons of cargo per week?
D. Assume that 85% of the time the port can handle the weekly cargo volume without extending operating hours. What is the number of tons of cargo per week that will require the port to extend its operating hours?

(A) Probability that the port handles less than 5 million tons of cargo per week is 0.7291.

(B) Probability that the port handles 3 or more million tons of cargo per week is 0.9664.

(C) Probability that the port handles between 3 million and 4 million tons of cargo per week is 0.2373.

(D) The number of tons of cargo per week that will require the port to extend its operating hours is 3.65 million tons.

Step-by-step explanation:

We are given that the port of South Louisiana, located along 54 miles of the Mississippi River between New Orleans and Baton Rouge.

The U.S. Army Corps of Engineers reports that the port handles a mean of 4.5 million tons of cargo per week with a standard deviation of 0.82 million tons.

Let X = Number of tons of cargo handled per week

SO, X ~ Normal()

The z score probability distribution for normal distribution is given by;

                               Z  =    ~ N(0,1)

where, = population mean = 4.5 million tons

            = standard deviation = 0.82 million tons

(A) Probability that the port handles less than 5 million tons of cargo per week is given by = P(X < 5 million tons)

        P(X < 5) = P( < ) = P(Z < 0.61) = 0.7291

The above probability is calculated by looking at the value of x = 0.61 in the z table which has an area of 0.7291.

(B) Probability that the port handles 3 or more million tons of cargo per week is given by = P(X 3 million tons)

        P(X 3) = P( ) = P(Z -1.83) = P(Z 1.83)

                                                                          = 0.9664

The above probability is calculated by looking at the value of x = 1.83 in the z table which has an area of 0.9664.

(C) Probability that the port handles between 3 million and 4 million tons of cargo per week is given by = P(3 million tons < X < 4 million tons)

     P(3 million tons < X < 4 million tons) = P(X < 4) - P(X 3)

      P(X < 4) = P( < ) = P(Z < -0.61) = 1 - P(Z 0.61)

                                                = 1 - 0.7291 = 0.2709

      P(X 3) = P( ) = P(Z -1.83) = 1 - P(Z < 1.83)

                                                 = 1 - 0.9664 = 0.0336

The above probability is calculated by looking at the value of x = 0.61 and x = 1.83 in the z table which has an area of 0.7291 and 0.9664 respectively.

Therefore, P(3 < X < 4) = 0.2709 - 0.0336 = 0.2373

(D) Now, it is given that 85% of the time the port can handle the weekly cargo volume without extending operating hours, that means;

           P(X > x) = 0.85     {where x is the required no. of tons of cargo}

           P( > ) = 0.85

           P(Z > ) = 0.85

In the z table, the critical value of x which represents top 85% area is given as -1.036, that is;

                     x  =  4.5 - 0.85 = 3.65 million tons

Hence, the number of tons of cargo per week that will require the port to extend its operating hours is 3.65 million tons.

answer: i'm not sure

step-by-step explanation: